20190908, 17:45  #1 
Sep 2002
Database er0rr
2^{5}·109 Posts 
A Flawed Proof of Fermat's Last Theorem
I just put the attached paper together. Comments will be welcomed.
Last fiddled with by paulunderwood on 20190908 at 22:45 
20190908, 20:13  #2 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
3·37·43 Posts 
Introduction: "Some 350 years ago", can't you be more accurate?
Introduction: Take out the double "famous"....this is not a contest who Where's the list of references?! Edit: Need to go to bed, will read the rest on my return from A/L. Last fiddled with by pinhodecarlos on 20190908 at 20:22 
20190908, 20:46  #3  
Jun 2015
Vallejo, CA/.
5×193 Posts 
Quote:
I just wonder if this might have been the "proof" our friend Pierre found that it would not fit in the margin of the book Arithmetic by Diophantus? Now THAT was a famous mathematician. But of course, all this is completely veering off the subject. 

20190908, 22:47  #4 
Sep 2002
Database er0rr
3488_{10} Posts 
Thanks for your input. I have replaced the original document (see above) to include your suggestions and several corrected mistakes.
Last fiddled with by paulunderwood on 20190908 at 22:49 
20190909, 08:11  #5 
Dec 2012
The Netherlands
2×3^{2}×83 Posts 
I don't yet understand what you are trying to accomplish here.
(Once you assume something that is not actually true, anything you like can then be proved by contradiction, of course.) Are you saying: "Fermat thought he had a proof and this may be what he had in mind"? Is the assumption that F_n is prime for all n supposed to be the only flaw or is the paper supposed to be full of flaws? 
20190909, 09:14  #6  
Sep 2002
Database er0rr
2^{5}·109 Posts 
Quote:
I think F_n is prime is the only flaw, but I might be wrong! 

20190909, 18:12  #7 
Sep 2002
Database er0rr
3488_{10} Posts 
I made a goof and so can simplify things greatly.
\(b^{2p}+c^{2p}\) can only be divisible by primes of the form \(4k+1\) (and maybe \(p\)). Hence there is no need to consider \(7,11,\ldots\) etc, With this in mind, consider \(97 \mid U_p\) as an example. Then \(a^{8\cdot p \cdot 6} + b^{8\cdot p \cdot 6} \equiv 0 \pmod{97}\). But the unique "base" solution is \(9^2+4^2 \equiv 0 \pmod{97}\). Can I say \(9^4\) should be equivalent to unity? It is not, it is \(62\). Last fiddled with by paulunderwood on 20190909 at 19:10 
20190909, 20:02  #8 
Jun 2015
Vallejo, CA/.
965_{10} Posts 
I am not claiming I have the full understatement of these issues.
But for instance, there are literally hundreds of conditional proofs assuming the RH (Riemann Hypothesis) is true. A few of those proofs have been perfected and are no longer dependent on the RH to be true. In other words there are true independently of whether the RH is true or false. (As an aside, this is considered some sort of validation of the RH being true. Not proof itself but evidence in favor.) Now assume there is a conjecture lets call it the Steele, Richardson Conjecture that is considered valid if the RH is proven. Now furtherly assume that for some quirk or extraordinary breakthrough the RH is PROVEN FALSE. Does this automatically mean The SteeleRichardson Conjecture is False? I would contend that it does not. It only means that the conditional proof arrived at is no longer a valid path for proving the SR conjecture. I do apologize for not using the exact mathematical language or terminology as I am just a dilettante, not a mathematician. Last fiddled with by rudy235 on 20190909 at 21:00 Reason: wether, weather, whether! 
20190910, 00:14  #9 
Sep 2002
Database er0rr
2^{5}·109 Posts 
Is there any easy way to find the solutions to Fermat's prime as the sum of two squares other than by brute force trial and error?

20190910, 01:24  #10  
"Curtis"
Feb 2005
Riverside, CA
1172_{16} Posts 
Quote:
Consider the statement "if it snows, then it is cold outside." On the 5th of March, it did not snow. Was it cold outside? On the 10th of March, it was not cold outside. Did it snow? So, a proof that goes "If RH is true, then XYZ" says nothing about XYZ when RH is false. However, if you learn XYZ is false, then RH must be false. Last fiddled with by VBCurtis on 20190910 at 01:25 

20190910, 10:13  #11  
Dec 2012
The Netherlands
2×3^{2}×83 Posts 
Quote:


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